Answer
$\dfrac{x^2-6x+3}{(x-2)^3}=\dfrac{1}{x-2}-\dfrac{2}{(x-2)^2}-\dfrac{5}{(x-2)^3}$
Work Step by Step
Here, we have $\dfrac{x^2-6x+3}{(x-2)^3}=\dfrac{A}{x-2}+\dfrac{B}{(x-2)^2}+\dfrac{C}{(x-2)^3}$
$x^2-6x+3 =A(x-2)^2+B(x-2)+C$
or, $x^2-6x+3 =Ax^2+(-4A+B)x+4A-2B+C$
Equating the coefficients, we get
$A=1$ and $ -4A+B = -6 \implies B=-2$
and $4A-2B+C =3 \implies C=-5$
Hence, $\dfrac{x^2-6x+3}{(x-2)^3}=\dfrac{1}{x-2}-\dfrac{2}{(x-2)^2}-\dfrac{5}{(x-2)^3}$
